TRANSIENT HEAT CONDUCTION

The variation of temperature with time as well as position

in one- and multidimensional systems.

350C 200C 300C 200C

Q 2  Q1

30 min 60 min

Variation of temperature with time in one dimensional systems:

HIGH PRACTICE

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 TRANSIENT HEAT CONDUCTION

During the heating process, the changes in surface The heat transfer process
temperature and product center are not the same depends on the position
In case of small materials with high conductivity  inner temperatures: the same
TRANSIENT HEAT CONDUCTION
Principle for determining the distribution of temperature in
one dimensional T(x,t)

Properties of material (Internal factor)

Parameters Dimensional Unit

Initial temperature of a body Ti 0C

Temperature of a body at 0C
T(x,t)
time t
Thermal conductivity of body k W/m2.0C
Density of body  ρ  kg/m3
Heat capactiy Cp J/kg.0C
½ thickness, radius L, r m

The measured/caculated position X, r0 m

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TRANSIENT HEAT CONDUCTION
Principle for determining the distribution of temperature in
one dimensional T(x,t)

Properties of fluid (surrounding)

Parameters Dimensional Unit

Surrounding medium temperature T∞ 0C

Convection heat transfer

 
h W/m2.0C
coefficient

Other parameters

Parameters Dimensional Unit

Time
 t
 s
Thermal Diffusivity α m2/s

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TRANSIENT HEAT CONDUCTION

Problems in transient heat conduction

1 Estimate the time to heat (or cool) the material from

the initial temperature T0 to the desired temperature T.

2 Predict the product's center temperature after the

heating/cooling time

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 TRANSIENT HEAT CONDUCTION

Problems in transient heat conduction

Cooling of fruit, egg boiling Heating/

Pasteurization
Freezing

Dimensionless heat transfer

The internal/surface temp. of body coefficient, Biot number (Bi)
Depending
on Dimensionless distance
Heating/Cooling time to desired temp (F0)
from the center
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the proper form of the dimensionless time
TRANSIENT HEAT CONDUCTION

Problems in transient heat conduction

Properties of material Prevents

Different Bi number
heat transfer from outside to inside
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TRANSIENT HEAT CONDUCTION

Calculation of L: characteristic length Depend on position

In the central:
radius (1/2 Diametet)

In the surface
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 TRANSIENT HEAT CONDUCTION

Bi < 0,1 (Rt << Rn) 0,1 < Bi < 40

Thermal resistance of inner Rt < Outside Rn Both inner and outside
thermal resistance have effect
the variation of temperature with
location within the body is slight and Using heat transfer equation
can reasonably be approximated as
being uniform.

Bi > 40 (Rt >> Rn)

𝑅𝑡 ℎ𝐿 Thermal resistance of inner Rt > Outside Rn
𝐵𝑖 = =
𝑅𝑛 𝑘
Outsied Thermal resistance  0

Surface temperature of body T=

environment temp. (T)
TRANSIENT HEAT CONDUCTION

Bi < 0,1

𝑑𝑇
hAs [T∞ − T(t)] = mCp
𝑑𝑡

m= .V

t = 0: T = Ti
t= t: T = T(t)
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TRANSIENT HEAT CONDUCTION

𝑇 − 𝑇∞ −ℎ𝐴𝑠 𝑡 ℎ.𝐿
= exp −ℎ𝑡 −( 𝑐 )𝑡
𝑘
𝑇𝑖 − 𝑇∞ 𝜌𝐶𝑝 𝑉 = exp 𝜌.𝐶 .𝐿 = exp (𝜌.𝐶 .𝐿 .𝐿
𝑝 𝑐 𝑝/𝑘) 𝑐 𝑐

𝛼𝑡
Fourier number 𝐹0 = 2
𝐿
The proper form of the dimensionless time (F0)
ℎ𝐿
𝐵𝑖 =
𝛼𝑡 (Bi) Biot number 𝑘
𝑇 − 𝑇∞ 𝑡
= exp( − 𝐵𝑖. )= exp( − 𝐵𝑖. 2 )
𝑇𝑖 − 𝑇∞ 1ൗ . 𝐿2 𝐿𝑐
𝛼 𝑐

𝑇 − 𝑇∞
= exp( − 𝐵𝑖. 𝐹0 )
𝑇𝑖 − 𝑇∞
TRANSIENT HEAT CONDUCTION

0, 1< Bi < 40

Dimensionless
Fourier number Biot number temperature  (x,t)
𝛼𝑡 ℎ𝐿 𝑇 − 𝑇∞
𝐹0 =  = 𝐵𝑖 = 𝜃(𝑥, 𝑡) =
𝐿2 𝑘 𝑇𝑖 − 𝑇∞

Approximate Analytical and Graphical Solutions

Method 1: Using Transient temperature and heat transfer charts
Method 2: Using tables (in case of F0 > 2)
 Determine uniform temperature of body
 Determine heating/cooling time

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TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40

Problem 1: Determine of
internal/surface
temperature of body
Calculate of Bi

Fourier number

From chart  determine

Dimensionless temp. (x,t)

T = T∞ + (x,t)[T-T∞]
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40

Problem 2: Determine
heating/cooling time

Find Bi number

Dimensionless temp.
(x,t)
From the chart,
determine F0

t = F0.L2/α (s)
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40
TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40
TRANSIENT HEAT CONDUCTION
Plate/walls Cylinder Sphere
0, 1< Bi < 40

With F0 > 2,
using equation

𝜃(𝑥, 𝑡) = 𝐶1 exp( − 𝜁 2 . 𝐹0 )

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TRANSIENT HEAT CONDUCTION
0, 1< Bi < 40

Plate/walls Cylinder Sphere

With F0 > 2,
using equation

𝜃(𝑥, 𝑡) = 𝐶1 exp( − 𝜁 2 . 𝐹0 )

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TRANSIENT HEAT CONDUCTION

𝑇 − 𝑇∞ 𝐿
= 𝑒𝑟𝑓
𝑇𝑖 − 𝑇∞ 2 𝛼𝑡
Surface temperature of will increase rapidly to the
ambient temperature
𝐿
𝑥=
2 𝛼𝑡

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TRANSIENT HEAT CONDUCTION

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TRANSIENT HEAT CONDUCTION

Dimenional
Mutidimensional
heat transfer
3D
Q
Q
Rectangular
with high infinite

Q Q Q Q

2D
Limited cylindrical Q
geometry
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TRANSIENT HEAT CONDUCTION

3D General principle

𝑇 − 𝑇∞ Bi ≠ depend on different L of each side

Side X
𝑇𝑖 − 𝑇∞ 𝑋 Different Dimensionless temp.

𝑇 − 𝑇∞ Calculate for each side

Side Y
𝑇𝑖 − 𝑇∞ 𝑌

𝑇 − 𝑇∞ Determine overal
Side Z dimenstionless temperature
𝑇𝑖 − 𝑇∞ 𝑍

𝑇 − 𝑇∞ 𝑇 − 𝑇∞ 𝑇 − 𝑇∞ 𝑇 − 𝑇∞
= × ×
𝑇𝑖 − 𝑇∞ 𝑇𝑖 − 𝑇∞ 𝑋
𝑇𝑖 − 𝑇∞ 𝑌
𝑇𝑖 − 𝑇∞ 𝑍

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Temperature Measurement by Thermocouples
The temperature of a gas stream is to be
measured by a thermocouple whose junction can
be approximated as a 1-mm-diameter sphere, as
shown in Fig. 11–9. The properties of the junction
are k = 35 W/moC, = 8500 kg/m3, and Cp = 320
J/kgoC, and the convection heat transfer
coefficient between the junction and the gas is h =
210 W/m2·oC. Determine how long it will take for
the thermocouple to read 99 percent of the initial
temperature difference.

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Temperature Measurement by Thermocouples

when Ti = 0C and T = 100C

a thermocouple is considered to have read 99%
of this applied temperature difference when its
reading indicates T(t) = 99C

𝑇 − 𝑇∞ −ℎ𝑡 h = 210 W/m2oC;  = 8500 kg/m3, Time t =

= exp Cp = 320 J/kgoC, Lc =1.67x10-4 m 10s
𝑇𝑖 − 𝑇∞ 𝜌. 𝐶𝑝 . 𝐿𝑐
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 Exercises
Predicting the Time of Death
A person is found dead at 5PM in a room whose temperature is 20C. The
temperature of the body is measured to be 25C when found, and the heat
transfer coefficient is estimated to be h=8 W/m2.C. Modeling the body
as a 30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of
that person.

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Predicting the Time of Death: A person is found dead at 5PM in a room whose
temperature is 20C. The temperature of the body is measured to be 25C when found,
and the heat transfer coefficient is estimated to be h=8 W/m2.C. Modeling the body
as a 30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of that person.

T (C)  Cp k
30 996 4178 0.615
31 996 4178 0.617
35 994 4178 0.623

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Predicting the Time of Death: A person is found dead at 5PM in a room whose
temperature is 20C. The temperature of the body is measured to be 25C when found,
and the heat transfer coefficient is estimated to be h=8 W/m2.C. Modeling the body
as a 30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of that person.

Problem 2: Determine
heating/cooling time
𝑇 − 𝑇∞ 25 − 20
𝜃(𝑥, 𝑡) = = = 0.3
𝑇𝑖 − 𝑇∞ 37 − 20 Find Bi number
1 1 Fo =1.05
= = 1.1236
𝐵𝑖 0.89 Dimensionless temp.
(x,t)
The time of death of that person
From the chart,
t = F0.L2/α with α = k/.Cp = 0.617/(996*4178)
determine F0
= 1.4827.10-7 m2/s
t = 1.05*0.06892/1.4827.10-7 = 33618 s = 9.34 h
t = F0.L2/α (s)

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33
Predicting the Time of Death: A person is found dead at 5PM in a room whose
temperature is 20C. The temperature of the body is measured to be 25C when found,
and the heat transfer coefficient is estimated to be h=8 W/m2.C. Modeling the body
as a 30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of that person.

Problem 2: Determine
heating/cooling time
𝑇 − 𝑇∞ 25 − 20
𝜃(𝑥, 𝑡) = = = 0.3
𝑇𝑖 − 𝑇∞ 37 − 20 Find Bi number
Fo =1.05
If using the equation
Dimensionless temp.
𝑇 − 𝑇∞ −ℎ𝑡
= exp (x,t)
𝑇𝑖 − 𝑇∞ 𝜌. 𝐶𝑝 . 𝐿𝑐
From the chart,
The time of death of that person determine F0
−8 ∗ 𝑡 t = 43150 s
0.3 = exp t = F0.L2/α (s)
996 ∗ 4178 ∗ 0.0689 t = 11.98 h
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Exercises
Boiling Eggs
An ordinary egg can be approximated as a 5-cm-diameter sphere. The egg is
initially at a uniform temperature of 5C and is dropped into boiling water at
95C. Taking the convection heat transfer coefficient to be h = 1200 W/m2·C,
determine how long it will take for the center of the egg to reach 70C.

Heating of Brass Plates in an Oven

In a production facility, large brass plates of 4-cm

thickness that are initially at a uniform temperature of
20C are heated by passing them through an oven
that is maintained at 500 C. The plates remain in
the oven for a period of 7 min. Taking the combined
convection and radiation heat transfer coefficient to
be h =120 W/m2C determine the surface
temperature of the plates when they come out of the
oven.

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Exercises
Cooling of a Short Brass Cylinder

A short brass cylinder of diameter D =10 cm and height H = 12 cm is

initially at a uniform temperature Ti =120C. The cylinder is now placed in
atmospheric air at 25C, where heat transfer takes place by convection,
with a heat transfer coefficient of h = 60 W/m2·C.
Calculate the temperature at (a) the center of the cylinder and (b) the
center of the top surface of the cylinder 15 min after the start of the
cooling.
Cooling a sphere
A person puts a few apples into the freezer at 15C to cool them quickly for guests
who are about to arrive. Initially, the apples are at a uniform temperature of 20C,
and the heat transfer coefficient on the surfaces is 8 W/m2C. Treating the apples as
9-cm-diameter spheres and taking their properties to be = 840 kg/m3, cp = 3.81
kJ/kgC, k = 0.418 W/mC, and a =1.3 107 m2/s, determine the center and surface
temperatures of the apples in 1 h. Also, determine the amount of heat transfer
from each apple.
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